Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void do_xor(int* a, int x, int y, vector<pair<int, int> >& v) {
  a[x] ^= a[y];
  v.emplace_back(x, y);
}
void do_swap(int* a, int x, int y, vector<pair<int, int> >& v) {
  do_xor(a, x, y, v);
  do_xor(a, y, x, v);
  do_xor(a, x, y, v);
}
int n;
vector<pair<int, int> > solve(int* a) {
  vector<pair<int, int> > ret;
  for (int i = 29, j = 1, k; i >= 0; --i) {
    for (k = j; k <= n && !(a[k] >> i & 1); ++k)
      ;
    if (k > n) continue;
    if (k != j) do_swap(a, k, j, ret);
    for (k = 1; k <= n; ++k)
      if ((a[k] >> i & 1) && (k != j)) do_xor(a, k, j, ret);
    ++j;
  }
  return ret;
}
const int N = 10010;
int a[N], b[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) cin >> b[i];
  auto u = solve(a), v = solve(b);
  for (int i = 1, j, k = 29; i <= n && a[i]; ++i) {
    for (; !(a[i] >> k & 1); --k)
      ;
    for (j = 1; j <= 30 && j <= n; ++j)
      if ((a[j] ^ b[j]) >> k & 1) do_xor(a, j, i, u);
  }
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i]) {
      cout << "-1\n";
      return 0;
    }
  cout << (u.size() + v.size()) << '\n';
  reverse(v.begin(), v.end());
  for (auto i : u) cout << i.first << ' ' << i.second << '\n';
  for (auto i : v) cout << i.first << ' ' << i.second << '\n';
  return 0;
}